1
2
3
4
5
6
7
δ0 a
a
Ab0
0
d d
d
a
0
δ
Ab0
0
d
a
a
b0 b
a
a
b0 b
a
8
M Tu
= ∑ γ u ,i ×M T ,i = 1.6 × M Tg + 1.8 × M Tp
δ0
i
b0 = b − 2 a ; d 0 = d − 2 a
a
0
δ
Ab 0 = b0 × d 0
Ab0
0
d
Ob0 = 2 × (b0 + d 0 )
δ 0 =
d m 8
τn =
=
min .(b0 , d 0 ) 8
M Tu 2 × Ab0 × δ0
a
> τr < 5 τ r
a
b0 b
a
9
10
τn ≤ 3 τr ⇒ M Tbu
M Tbu
=
1 × (3 τr − τn ) × 2 × Ab0 × δ0 2
= (3 τr − τn ) × Ab0 × δ0 ( 1 ) u
a
∑ A
M TRu
= M Tu − M Tbu
=
M TRu × tg θ × eu 2 × Ab0 × σv
=
M Tu × ctg θ × Ob0 2 × Ab0 × σv
a
( 1 ) p
a
=
M Tu × ctg θ × e p 2 × Ab0 × σv
11
12
τn = τRu
τn ≥ 3 τr ⇒
M Tbu M TRu
τn τRu
= 0 = M Tu − M Tbu = M Tu
13
Prim er 1: 1 : r e d it i t i n o m in i n al a l n i n ap a p o n s m ic i c an a n ja j a n i p o Ø O d re p o t r eb e b i i zv zv r ši ti t i o s i g u r an an j e ar m at u r o m p r av a v o u g ao a o n o g p r es e s ek e k a o p te t e r e enog e nog m o m e n ti t i m a t o r zi z i j e M T g i M T p u s le l e d s ta t al n o g , o d n o sn s n o p o vr v r e m en e n o g o p te t e r e e n j a. a . P o d ac ac i za p r o r a # u n : #
Ø M T g = 4 0 k N m b = 5 0 c m MB 30 N m d = 6 0 c m G A 24 2 4 0 /3 6 0 Ø M T p = 2 5 k Nm
M Tu
= 1.6 × 40 + 1.8 × 25 = 109 kNm
b0 = b − 2 a = 50 − 2 × 4
= 42 cm
d 0 = d − 2 a = 60 − 2 × 4
= 52 cm
Ab 0 = b0 × d 0 = 42 × 52 = 2184 cm 2 min .(b0 , d 0 ) 42 = = 5 .25 cm 8 8
δ0 =
d m 8
τn =
109 ×10 2 kN = 0 .475 2 2 × 2184 × 5 .25 cm
=
> τ = 0 .11 kN r cm 2 < 5 τr
14
τn > 3 τr ⇒
= 0
; M TRu
=
M TRu × tg θ × eu 2 × Ab 0 × σv
=
109 ×10 2 ×1.0 × eu 2 × 2184 × 24
( 1 ) u
a
au( 1 )
M Tbu
UØ8
⇒
eu
=
UØ10
⇒
eu
=
UØ12
⇒
eu
=
= M Tu − M Tbu = M Tu
15
= 0 .104 × eu
0 .503 = 4.83 cm 0 .104 0 .785 = 7 .55 cm ⇒ usv . UØ10 / 7 .5 cm 0 .104 1.13 = 10 .88 cm 0 .104
∑ A
a
=
M Tu × ctg θ × Ob0 2 × Ab 0 × σv
16
Ob0 = 2 × (b0 + d 0 ) = 2 × (42 + 52 ) = 188 cm
∑ A
a
=
109 ×10 2 ×1.0 ×188 = 19.55 cm 2 2 × 2184 × 24
Smanjenje ugla θ 109 × 10 2 a = × tg 35 ° × eu 2 × 2184 × 24 0 .785 ( 1 ) au = 0 .073 × eu ⇒ eu = = 10 .79 cm ⇒ usv . UØ10/10 0 .073 2 109 × 10 A a = × ctg 35 ° ×188 = 27 .92 cm 2 ⇒ usv .14Ø16 2 × 2184 × 24 ( 1 ) u
∑
2Ø14
2Ø14 2Ø14 UØ10/7.5 2Ø14
4Ø14
17
18
Prim er 2: 2 : Za nosa POS1 prikazan na skici nacrtati dijagrame presenih sila usled prikazanih sila G=75 kN, a zatim izvršiti osiguranje od glavnih napona zatezanja na pojedinim delovima nosaa. Presek nosaa je pravougaoni, dimenzija b/d = 45/60 cm. Nosa nije potrebno dimenzionisati prema momentima savijanja. Kvalitet materijala: MB 30, GA 240/360.
19
K r aj a j n je j e t r e i n e n o sa s a # a # : T u = 1.6 × ×75 = 120 k N pr etp. a 6 0 - 5 = 5 5 c m ⇒ 1 = 5 cm ⇒ h = 60 2 (T)= (T) = 0.054 0. 054 k N/cm n
M T u = 1.6 × ×25 = 40 kNm pr etp. a 4.5 cm ⇒ b 0 = 4 5 - 2 × 4.5 4. × 5 = 36 c m ; 0 = 4.5 d 0 = 6 0 - 2 × 4.5 × = 51 c m = 4.5 4.5 c m
20
21
A b 0 = b 0 × d 0 = 3 6 × ×51 = 1836 cm 2 ; O b 0 = 2 × ) = 174 c m ×( b 0 + d × ×(36+51) 0 = 2 M T τn
τn
=
=
M Tu 2 × Ab 0 × δ
M T τn
+
T τn
=
40 × 10
2
2 × 1836 × 4.5
= 0.242kN / cm
= 0.242 + 0.054 = 0.296kN / cm
2
2
P o t r eb e b n e u ze z e n g i je j e z a p ri r i h v at a t an a n j e t ra r a n s v er e r za z a l n e s i le le: T 1 τn
Tbu
= ×
1
0.054
2
×
2
τn
0.296
× ( 3 × τ r − τ n ) × b × z =
× ( 3 × 0.11 − 0.296 ) × 45 × 0.9 × 55 = 6.9 kN
22
T Ru T τ Ru
= 120 − 6.9 = 113.1kN =
113.1 4 5 × 0. 9 × 5 5
u s v o j en o : (1) u ,T
a
=
=
= 0.051kN / cm 2 = 9 0 ° ; θ = 45°
m =4 ; T b × τ Ru
m ×σ v
45 × 0.051 4 × 24
× ×
1
( cos α + sin α × ctgθ ) 1
(1 + 0 ×1)
× eu =
× eu = 0.024 ⋅ eu
23
P o t r eb e b n e u ze z e n g i je j e z a p ri r i h v at a t an a n j e t o rz r z i je je
M Tbu
=
=
M T
τn
τn
0.242 0.296
× ( 3 ×τ r − τ n ) × Ab 0 × δ =
× ( 3 × 0.11 − 0.296 ) ×1836 × 4.5 ×10−2 = 2.3kNm
M TRu = M Tu - M Tbu = 40 - 2.3 = 37.7 kNm (1) u ,M T
a
=
=
M TRu 2 × Ab 0 × σ v
37.7 ×10 2 2 × 1836 × 24
× tgθ × eu =
× 1× eu = 0.043 ⋅ eu
24
sp olj ašnje uzengije: ) au(1, spolja = au(1, M ) T
+ au(1,T ) = (0.043 + 0.024) ⋅ eu = 0.067 ⋅ eu
pretp. U Ø10 (a u(1) = 0.785 cm2 ) eu ,spolja =
0.785 0.067
⇒
= 11.8cm → U Φ10 / 10
un utr ašnje uzeng ije: ) au(1,unutra = au(1,T )
= 0.024 ⋅ eu
pretp. U Ø8 (a u(1) = 0.503 cm2 )
eu ,unutra =
0.503 0.024
⇒
= 21.1cm → U Φ8 / 20
25
H o r iz i z o n t al a l n a ar m a t u r a
∑ ∆
a
a
=
=
M Tu 2 × Ab 0 × σ v
T mu 2 ×σ v
× ctgθ × Ob 0 =
× ( ctgθ − ctgα ) =
40 × 102 2 × 1836 × 24
120 2 × 24
× 1× 174 = 7.9cm2
× (1 − 0) = 2.5cm2
26
Prim er 3: 3 : # Ø D i m en e n zi z i o n is i s at a t i n o s a # si st em a ob o st ran o uk lještene gr ede, rasp on a L=6.0 L=6.0 m . Pored so ps tvene te ine, # en e n jednako n o s a # je j e o p t e r e jedna ko raspodeljenim raspodelje nim stalnim stal nim enjem e njem , kao i ∆g) ( ∆ g ) i po vrem enim (p) op tere raspodeljenim raspodelje nim m om e ntim a torzije torz ije m T g i m T p usled nja. s t al a l n o g , o d n o s n o p o v r e m e n o g o p t er e r e enja. e #no # o g p reseka us vo jiti iz Dimenzije po pre n i z us lov a da je napon sm icanja usled m om enata ena ta torzije torzije n(MT) = 2 r . #u # n : Podaci za pro ra Ø m Tg = 6 kNm / m ∆ g = 1 6 .5 k N /m M B 30 Ø m Tp = 4 k N m / m p = 4 k N /m G A 2 4 0 /3 6 0
g, p
27
mTg, mTp
p o
M
M o
M
M
T T
MT T
M
T
T
M
6 .0 × 6 .0 M Tg = = 18 kNm ; M Tp 2 M Tu
τ
M T n
=
4.0 × 6 .0 = 12 kNm 2
= 1.6 ×18 + 1.8 ×12 = 50 .4 kNm =
M Tu 2 × Ab0 × δ0
= 0 .22 kN / cm 2 = 2 τr
Ab0 = b0 × d 0 = (b − 2 a ) × (d − 2 a )
δ0 =
d m 8
=
min .(b0 , d 0 ) min .(b − 2 a, d − 2 a ) = 8 8
pretpostavljeno: a = 4 cm, b d
28
Ab0 = (b − 2 × 4 ) × (d − 2 × 4 ) = (b − 8 ) × (d − 8 )
δ0 =
b − 8 8
⇒
50 .4 ×10 2 b − 8 2 × (b − 8 ) × (d − 8 ) × 8
2
(b − 8 ) × (d − 8 ) =
8 × 50 .4 ×10 2 2 × 0 .22
= 0 .22
= 91636 .4 cm 3
b
d
usv . d
1
40
97.5
100
2
45
74.9
75
3
50
59.9
60
4
55
49.5
< b !
29
30
- sopstvena težina 0.50 ×0.60 ×25 Dg ukupno, stalno optere ! enje: g
M g o
p g
M
=
24.0 × 6 .0 2 12
=
24.0 × 6 .0 24
T g =
2
= 72 .0 kNm = 36 .0 kNm
24.0 × 6 .0 = 72 .0 kN 2
o p
M
M p p
T p
= 7.5 kN/m = 16.5 kN/m = 24.0 kN/m
=
4.0 × 6 .0 2 12
= 12 .0 kNm
=
4.0 × 6 .0 2 24
= 6 .0 kNm
=
4.0 × 6 .0 = 12 .0 kN 2
31
oslonac
= 1.6 × 72 + 1.8 ×18 = 136 .8 kNm 50 × 55 .5 2 .05 Aa , potr . = 4.490 × × = 10 .64 cm 2 M u
100
24
polje M u
= 1.6 × 36 + 1.8 × 9 = 68 .4 kNm
b + 8 × d p = 50 + 8 × 12 = 146 cm B = min . 0 .25 0 .25 = 85 cm + × = + × × = b l 50 0 . 7 600 85 cm 0 3 3
Aa , potr .
85 × 55 .5 2 .05 = 1.424 × × = 5 .48 cm 2 100 24
Aa ,min .
b × d 50 × 60 = µ min × = 0 .25 × = 7 .5 cm 2 100 100
z ≈ 0 .9 × h = 0 .9 × 55 .5 = 50 cm T u
= 1.6 × 72 + 1.8 ×18 = 136 .8 kN
τ = T n
T mu 136 .8 = = 0 .055 kN / cm 2 b × z 50 × 50
Ab0 = b0 × d 0 = 42 × 52 = 2184 cm 2
δ0 = τM = n T
d m 8
42 = = 5 .25 cm 8 M Tu 50 .4 ×10 2 = = 0 .22 kN / cm 2 2 × Ab0 × δ0 2 × 2184 × 5 .25
32
33
τn = τ + τ T n
λ=
M T n
= 0 .055 + 0 .220 = 0 .275 kN / cm
> τr = 1.1 MPa < 5 τ r
L τr 600 1.1 × 1 − = × 1 − = 180 cm 2 τn 2 2 .75
τn = 2 .75 MPa < 3 τr = 3.3 MPa ⇒ T bu T bu
2
=
=
T bu
> 0 ;
M Tbu
> 0
1 τT n × × (3 τr − τn ) × b × z 2 τn
1 0 .055 × × (3 × 0 .11 − 0 .275 ) × 50 × 50 = 13.8 kN 2 0 .275
34
T Ru
= 136 .8 − 13.8 = 123.0 kN
τ = T Ru
123.0 kN = 0 .049 2 50 × 50 cm
usvojeno : m = 4 ; ( 1 ) u ,T
a
( 1 ) u ,T
a
α = 90 °
;
θ = 45 °
b × τT Ru
=
1 × × eu m × σv (cos α + sin α × ctg θ )
=
50 × 0 .049 1 × × eu (0 + 1 ×1) 4 × 24
= 0 .026 × eu
M Tbu M Tbu
τ = × (3 τr − τn ) × Ab0 × δ0 τn
=
M TRu ( 1 ) u ,M T
a
au( 1, )M T
M T n
35
0 .22 × (3 × 0 .11 − 0 .275 ) × 2184 × 5 .25 ×10 −2 = 5 .09 kNm 0 .275
= 50 .4 − 5 .09 = 45 .31 kNm =
M TRu × tg θ × eu 2 × Ab0 × σv
=
45 .31 × 10 2 ×1.0 × eu 2 × 2184 × 24
= 0 .043 × eu
36
au( 1, )spolja
= au( 1, )M + au( 1,T ) = (0 .043 + 0 .026 ) × eu = 0 .069 × eu T
UØ10 ⇒ eu ,spolja
au( 1, )unutra
=
0 .785 = 11.4 cm ⇒ usv . UØ10/10 0 .069
= au( 1,T ) = 0 .026 × eu
UØ8 ⇒ eu ,unutra
=
0 .503 = 19.6 cm ⇒ usv . UØ8/20 0 .026
∑ A = a
37
M Tu × ctg θ × Ob0 2 × Ab 0 × σv
Ob 0 = 2 × (b0 + d 0 ) = 2 × (42 + 52 ) = 188 cm
∑ A
a
=
50 .4 × 10 2 ×1.0 ×188 = 7 .90 cm 2 ⇒ usv .14Ø10 2 × 2184 × 24 7Ø16
∆ Aa = 0 (" špic " momenta )
2Ø10 2Ø10
za M T : za M: ukupno: usvojeno:
4Ø10 = 3. 3.14 cm 2 = 10.64 cm 2 = 13.78 cm 2 (14.07 cm 2 ) 7Ø16 (14.07
UØ8/20 UØ10/10 2Ø10
4Ø10
